Question

Fluoride ion is poisonous in relatively low amounts: 0.2 g of F - per 70 kg body weight can cause death. Nevertheless, in order to prevent tooth decay, F - ions are added to drinking water at a concentration of 1 mg of F - ion per L of water. How many liters of fluoridated drinking water would a 70 kg person have to consume in one day to reach this toxic level?

L
How many kilograms of sodium fluoride would be needed to treat a 7.75 107 gal reservoir?
kg

Answer 1

200 liters of fluoridated drinking water

293 kilograms of sodium fluoride

Step-by-step explanation:

0.2g of fluoride ion per 70kg body weight can cause death

1mg (0.001g) of fluoride ion per liter of water prevents tooth decay

Liter of fluoridated drinking water a 70kg person would have to consume to reach the toxic level (0.2g) = 0.2g × 1L/0.001g = 200L

Volume of reservoir = 7.75×10^7gal = 7.75×10^7 × 3.7854L = 2.93×10^8L

1mg (10^-6kg) of sodium fluoride is needed to treat 1L of reservoir

Mass of sodium fluoride needed to treat 2.93×10^8L of reservoir = 2.93×10^8L × 10^-6kg/1L = 293kg